Problem: $\sum\limits_{n=1}^{\infty }\dfrac{(x-3)^n}{n\cdot 2^n}$ What is the interval of convergence of the series? Choose 1 answer: Choose 1 answer: (Choice A) A $1\leq x \leq5$ (Choice B) B $-2\leq x <2$ (Choice C) C $-2< x <2$ (Choice D) D $1\leq x <5$
Explanation: We use the ratio test. For $x\neq3$, let $a_n=\dfrac{(x-3)^n}{n\cdot 2^n}$. $\lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|=\left|\dfrac{x-3}{2}\right|$ The series converges when $\left|\dfrac{x-3}{2}\right|<1$, which is when $1<x<5$. Now let's check the endpoints, $x=1$ and $x=5$. Letting $x=1$, we get the series: $\begin{aligned} \sum\limits_{n=1}^{\infty }\dfrac{(1-3)^n}{n\cdot 2^n}&=\sum\limits_{n=1}^{\infty }\dfrac{(-2)^n}{n\cdot 2^n} \\\\ &=\sum\limits_{n=1}^{\infty }\dfrac{(-1)^n\cdot 2^n}{n\cdot 2^n} \\\\ &=\sum\limits_{n=1}^{\infty }\dfrac{(-1)^n}{n} \end{aligned}$ This is the alternating harmonic series, which is known to converge. Letting $x=5$, we get the series $\begin{aligned} \sum\limits_{n=1}^{\infty }\dfrac{(5-3)^n}{n\cdot 2^n}&=\sum\limits_{n=1}^{\infty }\dfrac{2^n}{n\cdot 2^n} \\\\ &=\sum\limits_{n=1}^{\infty }\dfrac{1}{n} \end{aligned}$ This is the harmonic series, which is known to diverge. In conclusion, the interval of convergence is $1\leq x <5$.